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Dirac Lagr in Hermitian form

by ChrisVer
Tags: dirac, form, hermitian, lagr
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ChrisVer
#1
May11-14, 05:52 PM
P: 736
Is there any way to write the Dirac lagrangian to have symmetric derivatives (acting on both sides)? Of course someone can do that by trying to make the Lagrangian completely hermitian by adding the hermitian conjugate, and he'll get the same equations of motion (a 1/2 must exist in that case)...However same equations of motion, imply that there should be a connection between the two Lagrangians via a 4divergence, right? Unfortunately I cannot see what that could be.
[itex]L_{D}= \bar{\psi} (i γ^{\mu}∂_{\mu}-m) \psi [/itex]
In particular choosing
[itex]L_{D}= \bar{\psi} (i γ^{\mu}∂_{\mu}-m) \psi + ∂_{\mu}K^{\mu}[/itex]
seems to bring some result if I choose [itex]K^{\mu}=i\bar{\psi}\gamma^{\mu}\psi[/itex]
but the initial lagrangian seems to get doubled...
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ChrisVer
#2
May11-14, 06:11 PM
P: 736
[itex]L_{D}=\frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi + \frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi -m \bar{\psi}\psi[/itex]

[itex]L_{D}=\frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi + \frac{1}{2} i ∂_{\mu}(\bar{\psi}γ^{\mu}\psi) - \frac{1}{2}i (∂_{\mu}\bar{\psi})\gamma^{\mu}\psi -m \bar{\psi}\psi[/itex]

[itex]L_{D}=\frac{1}{2} \bar{\psi} (i γ^{\mu}[∂_{\mu}^{→}-∂_{\mu}^{←}] -2m) \psi + \frac{1}{2} i ∂_{\mu}(\bar{\psi}γ^{\mu}\psi)[/itex]

Should the - exist?
The_Duck
#3
May11-14, 06:23 PM
P: 841
Yes, your last expression is right.


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